Orienting Yourself: The Use of Coordinates — CBSE Class 9 Maths Textbook Solutions

From Fig. 1.3, the door $D_1R_1$D₁R₁ lies along the bottom wall of the room, which coincides with the $x$x-axis, so both $D_1$D₁ and $R_1$R₁ have $y=0$y=0 — the door is $0$0 ft from the $x$x-axis (it sits exactly on that wall). Reading the near end of the door, $D_1$D₁, off the grid gives $x=9.5$x=9.5, so the door begins $9.5$9.5 ft to the right of the $y$y-axis (left wall) — the door is $9.5$9.5 ft from the left wall.

Reading Fig. 1.3, $D_1$D₁ lies on the $x$x-axis, $9.5$9.5 units from the origin, so $D_1=(9.5,\,0)$D₁=(9.5, 0).

Since $D_1(9.5,0)$D₁(9.5,0) and $R_1(11.5,0)$R₁(11.5,0) lie on the same horizontal line ($y=0$y=0), the width of the door is simply $|11.5-9.5|=2$|11.5-9.5|=2 ft.

A width of $2$2 ft ($\approx 60$≈ 60 cm) is narrower than the standard comfortable interior door width of about $2.5$2.5–$3$3 ft ($75$75–$90$90 cm), so this is not a very comfortable width. A wheelchair typically needs a clear opening of at least about $2.7$2.7–$3$3 ft ($32$32–$36$36 inches); at only $2$2 ft wide, a person using a wheelchair would not be able to enter easily.

$B_1(0,1.5)$B₁(0,1.5) and $B_2(0,4)$B₂(0,4) both lie on the $y$y-axis, so the bathroom door's width is $|4-1.5|=2.5$|4-1.5|=2.5 ft.

Compared with the room door's width of $2$2 ft (found above), $2.5$2.5 ft $> 2$> 2 ft, so the bathroom door is wider than the room door, by $0.5$0.5 ft.

Typical interior room doors are about $2.5$2.5–$3$3 ft wide ($75$75–$90$90 cm, roughly $30$30–$36$36 inches); front/main entrance doors are often a little wider, around $3$3–$3.5$3.5 ft. Measuring actual doors at home and school (using a tape measure across the door itself, not just the frame) lets you compare real values against this typical range.

An accessible doorway generally needs a clear opening width of at least about $32$32–$36$36 inches ($0.8$0.8–$0.9$0.9 m, roughly $2.7$2.7–$3$3 ft). To check, measure the clear opening of each door (the gap available when the door is fully open, which is a little less than the door panel's own width because of the frame and hinges) and compare it against this $2.7$2.7–$3$3 ft benchmark — doors narrower than this would not comfortably admit a wheelchair.

Every point on the $y$y-axis has the form $(0,y)$(0,y), so its $x$x-coordinate is always $0$0.

Yes. Every point on the $x$x-axis has the form $(x,0)$(x,0), so its $y$y-coordinate is always $0$0.

Yes, but only in one case. $Q(y,x)=P(x,y)$Q(y,x)=P(x,y) requires the first coordinates to match, $y=x$y=x, and the second coordinates to match, $x=y$x=y — both conditions are the same equation, $x=y$x=y. So $Q$Q coincides with $P$P exactly when $x=y$x=y (then both points are simply $(x,x)$(x,x)); whenever $x\ne y$x≠ y, $Q$Q and $P$P are two different points.

Yes, the claim is true. Two ordered pairs are equal only when their corresponding entries match: $(x,y)=(y,x)$(x,y)=(y,x) needs $x=y$x=y (matching first entries) and $y=x$y=x (matching second entries) — both give exactly the condition $x=y$x=y. So $(x,y)=(y,x)$(x,y)=(y,x) if and only if $x=y$x=y; and whenever $x\ne y$x≠ y, the first entries $x$x and $y$y already differ, so $(x,y)\ne(y,x)$(x,y)≠(y,x).

(i) Label the three given feet $A(8,9)$A(8,9), $B(11,9)$B(11,9), $C(11,7)$C(11,7). $A\to B$A→ B is a horizontal move ($+3$+3 in $x$x) and $B\to C$B→ C is a vertical move ($-2$-2 in $y$y), so for $ABCD$ABCD to close up into a rectangle, the fourth foot $D$D must satisfy $D=A+(C-B)=(8,9)+(0,-2)=(8,7)$D=A+(C-B)=(8,9)+(0,-2)=(8,7). So the fourth foot is at $(8,\,7)$(8, 7).

(ii) This is a matter of judgement: the table occupies the region $8\le x\le 11$8≤ x≤ 11, $7\le y\le 9$7≤ y≤ 9, which (from Ex. 1.1) is well clear of the room door $D_1R_1$D₁R₁ (which sits along $y=0$y=0, between $x=9.5$x=9.5 and $x=11.5$x=11.5) and doesn't block the walking path from the door — so it is a reasonable, sensible spot, tucked along a wall rather than in the middle of the room.

(iii) Width $=|11-8|=3$=|11-8|=3 ft; length $=|9-7|=2$=|9-7|=2 ft. The height cannot be found from these coordinates — a floor-plan (top-down) view only records two dimensions (length and width); the height would need a side/elevation view, which isn't given here.

Using $B_1=(0,1.5)$B₁=(0,1.5) (from Ex. 1.1) as the hinge and the room door's width of $2.5$2.5 ft (the bathroom door's width found in Ex. 1.1): as the door swings open into the bedroom, it sweeps out a quarter-circle arc of radius $2.5$2.5 ft centred at $B_1$B₁. It will hit the wardrobe only if any part of the wardrobe lies within $2.5$2.5 ft of $B_1$B₁, inside that swing quadrant — check this using the distance formula between $B_1$B₁ and the wardrobe's nearest corner.

If the door is made wider, the swept radius increases, so a collision becomes more likely, not less. If there's already a tight clearance, the sensible fix is to move the wardrobe further from $B_1$B₁ (out of the swing arc), or re-hang the door so it opens the other way (e.g. outward, away from the wardrobe), rather than widening it.

These sub-parts are read directly off Fig. 1.5, so the working below uses one consistent, illustrative reading of the grid that matches the method you'd apply to the printed figure — check it against your own copy of Fig. 1.5.

(i) Since $O$O (the bathroom's corner) is the same point as the global origin, take the bathroom as the rectangle $O(0,0)$O(0,0), $F(6,0)$F(6,0), $R(6,8)$R(6,8), $P(0,8)$P(0,8) — a $6\text{ ft}\times 8\text{ ft}$6 ft× 8 ft room, with $OF$OF running along the $x$x-axis and $OP$OP along the $y$y-axis.

(ii) A corner shower unit $SHWR$SHWR with corners $(0,5)$(0,5), $(3,5)$(3,5), $(3,8)$(3,8), $(0,8)$(0,8): all four sides equal $3$3 ft and meet at right angles (two sides run parallel to the $x$x-axis, two parallel to the $y$y-axis), so $SHWR$SHWR is a square.

(iii) Washbasin ($3$3 ft $\times$× $2$2 ft): corners $(0,0)$(0,0), $(3,0)$(3,0), $(3,2)$(3,2), $(0,2)$(0,2). Toilet ($2$2 ft $\times$× $3$3 ft): corners $(3,0)$(3,0), $(5,0)$(5,0), $(5,3)$(5,3), $(3,3)$(3,3) — placed next to the washbasin so the two fittings don't overlap each other or the shower.

Again, exact placement depends on Fig. 1.5; the axes were set up to run from $x=-7$x=-7 to $13$13 and $y=-15$y=-15 to $12$12 precisely so that every room fits inside that boundary. A consistent way to fit an $18\text{ ft}\times 15\text{ ft}$18 ft× 15 ft dining room in that space (adjoining the bathroom/room drawn in positive coordinates) is:

(i) Corners $P(-7,-15)$P(-7,-15), $A(-7,3)$A(-7,3) (the length, $PA$PA, running vertically, $|3-(-15)|=18$|3-(-15)|=18 ft), and the other two corners at $(8,-15)$(8,-15) and $(8,3)$(8,3) (the width, $15$15 ft, since $|8-(-7)|=15$|8-(-7)|=15).

(ii) The centre of a rectangle is the midpoint of any pair of opposite corners: centre $=\left(\dfrac{-7+8}{2},\dfrac{-15+3}{2}\right)=(0.5,\,-6)$=((-7+8)/2,(-15+3)/2)=(0.5, -6). A $5\text{ ft}\times 3\text{ ft}$5 ft× 3 ft table centred here extends $2.5$2.5 ft either side in $x$x and $1.5$1.5 ft either side in $y$y, so its four feet are at $(0.5-2.5,\,-6-1.5)=(-2,-7.5)$(0.5-2.5, -6-1.5)=(-2,-7.5), $(0.5+2.5,\,-6-1.5)=(3,-7.5)$(0.5+2.5, -6-1.5)=(3,-7.5), $(3,\,-6+1.5)=(3,-4.5)$(3, -6+1.5)=(3,-4.5) and $(-2,-4.5)$(-2,-4.5).

Along the $x$x-axis: $|7-3|=4$|7-3|=4 units. Along the $y$y-axis: $|4-1|=3$|4-1|=3 units.

Yes. The horizontal shift ($4$4) and vertical shift ($3$3) are the two legs of a right triangle whose hypotenuse is $AD$AD, so by the Baudhāyana–Pythagoras theorem $AD=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$AD=√(4²+3²)=√(16+9)=√25=5 units.

Same: the side lengths $AD$AD, $DM$DM, $MA$MA (and so the size and shape of the triangle) — reflection preserves distances, so the triangle is congruent to its image. Changed: every point's $x$x-coordinate is negated (each point moves to the mirror-image position on the opposite side of the $y$y-axis) while its $y$y-coordinate is unchanged; the triangle's left–right orientation is reversed.

Yes, the same type of conclusion holds. Reflecting in the $x$x-axis is also an isometry, so side lengths and shape are again preserved; this time it is the $y$y-coordinates that get negated while the $x$x-coordinates stay the same (and the up–down orientation is reversed instead of left–right).

The two axes intersect at the origin, whose $x$x-coordinate is $0$0 and $y$y-coordinate is $0$0, i.e. the point $(0,0)$(0,0).

A line through $W$W parallel to the $y$y-axis is the vertical line $x=-5$x=-5, so any point $H$H on it has the same $x$x-coordinate as $W$W: $H=(-5,\,k)$H=(-5, k) for some value $k$k. Since the line $x=-5$x=-5 passes through Quadrant II (where $y>0$y>0), Quadrant III (where $y<0$y<0), and crosses the $x$x-axis itself (where $y=0$y=0), $H$H can lie in Quadrant II or Quadrant III (or on the $x$x-axis, if $k=0$k=0).

Join $R(3,0)\to A(0,-2)\to M(-5,-2)\to P(-5,2)\to R$R(3,0)→ A(0,-2)→ M(-5,-2)→ P(-5,2)→ R.

• Side $AM$AM: from $(0,-2)$(0,-2) to $(-5,-2)$(-5,-2) — same $y$y-coordinate throughout, so $AM$AM is horizontal, parallel to the $x$x-axis, length $5$5.
• Side $MP$MP: from $(-5,-2)$(-5,-2) to $(-5,2)$(-5,2) — same $x$x-coordinate throughout, so $MP$MP is vertical, parallel to the $y$y-axis, length $4$4.

(i) A horizontal side and a vertical side are always perpendicular, so $AM \perp MP$AM ⊥ MP.

(ii) Both $AM$AM (parallel to the $x$x-axis) and $MP$MP (parallel to the $y$y-axis) qualify; e.g. $AM$AM is parallel to the $x$x-axis.

(iii) $M(-5,-2)$M(-5,-2) and $P(-5,2)$P(-5,2) share the same $x$x-coordinate, and their $y$y-coordinates are negatives of each other ($-2$-2 and $2$2). Reflecting $M$M in the $x$x-axis (which negates $y$y, keeps $x$x) sends $(-5,-2)\to(-5,2)=P$(-5,-2)→(-5,2)=P. So $M$M and $P$P are mirror images of each other in the $x$x-axis.

Plotting confirms: $R$R is on the positive $x$x-axis, $A$A is on the negative $y$y-axis, $M$M is in Quadrant III, and $P$P is in Quadrant II.

One valid construction (many others are possible, as the book notes): choose $I=(5,0)$I=(5,0), directly above $Z$Z on the $x$x-axis, and $N=(0,-6)$N=(0,-6), directly left of $Z$Z on the $y$y-axis.

• $IZ$IZ: from $(5,0)$(5,0) to $(5,-6)$(5,-6) is vertical, length $=|0-(-6)|=6$=|0-(-6)|=6.
• $ZN$ZN: from $(5,-6)$(5,-6) to $(0,-6)$(0,-6) is horizontal, length $=|5-0|=5$=|5-0|=5.

Since $IZ$IZ is vertical and $ZN$ZN is horizontal, they are perpendicular, so the right angle is at $Z$Z — $IZN$IZN is right-angled with legs $IZ=6$IZ=6 and $ZN=5$ZN=5.

• $IN$IN (hypotenuse): $=\sqrt{(5-0)^2+(0-(-6))^2}=\sqrt{25+36}=\sqrt{61}\approx 7.81$=√((5-0)²+(0-(-6))²)=√(25+36)=√61≈ 7.81 units.

So the three sides are $6$6, $5$5, and $\sqrt{61}\approx 7.81$√61≈ 7.81 units.

Without negative numbers, both coordinates of every point would have to be $\ge 0$≥ 0, so we could only describe points lying in Quadrant I (including its boundary on the positive axes and the origin). No, such a system would not let us locate every point of the plane — points in Quadrants II, III and IV, and on the negative parts of the axes, would have no coordinates at all; it would only cover one quarter of the plane.

Method: compute the three pairwise distances and check whether the sum of the two shorter ones equals the longest.

$MA=\sqrt{(0-(-3))^2+(0-(-4))^2}=\sqrt{9+16}=\sqrt{25}=5$MA=√((0-(-3))²+(0-(-4))²)=√(9+16)=√25=5

$AG=\sqrt{(6-0)^2+(8-0)^2}=\sqrt{36+64}=\sqrt{100}=10$AG=√((6-0)²+(8-0)²)=√(36+64)=√100=10

$MG=\sqrt{(6-(-3))^2+(8-(-4))^2}=\sqrt{81+144}=\sqrt{225}=15$MG=√((6-(-3))²+(8-(-4))²)=√(81+144)=√225=15

Since $MA+AG=5+10=15=MG$MA+AG=5+10=15=MG, the points are collinear — yes, $M$M, $A$A, $G$G lie on the same straight line. (Equivalently: each point satisfies $y=\frac{4}{3}x$y=4/3x, since $-4/-3=4/3$-4/-3=4/3 and $8/6=4/3$8/6=4/3.)

$RB=\sqrt{(-2-(-5))^2+(-5-(-1))^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$RB=√((-2-(-5))²+(-5-(-1))²)=√(3²+(-4)²)=√(9+16)=√25=5

$BC=\sqrt{(4-(-2))^2+(-12-(-5))^2}=\sqrt{6^2+(-7)^2}=\sqrt{36+49}=\sqrt{85}\approx 9.220$BC=√((4-(-2))²+(-12-(-5))²)=√(6²+(-7)²)=√(36+49)=√85≈ 9.220

$RC=\sqrt{(4-(-5))^2+(-12-(-1))^2}=\sqrt{9^2+(-11)^2}=\sqrt{81+121}=\sqrt{202}\approx 14.213$RC=√((4-(-5))²+(-12-(-1))²)=√(9²+(-11)²)=√(81+121)=√202≈ 14.213

Check: $RB+BC=5+9.220=14.220$RB+BC=5+9.220=14.220, which is not equal to $RC\approx 14.213$RC≈ 14.213 (close, but not equal) — so $R$R, $B$B, $C$C are not collinear.

This is confirmed by comparing slopes: slope of $RB=\dfrac{-5-(-1)}{-2-(-5)}=\dfrac{-4}{3}$RB=(-5-(-1))/(-2-(-5))=-4/3, while slope of $BC=\dfrac{-12-(-5)}{4-(-2)}=\dfrac{-7}{6}$BC=(-12-(-5))/(4-(-2))=-7/6. Since $-\frac{4}{3}\ne-\frac{7}{6}$-4/3≠-7/6, the three points do not lie on one straight line — unlike the collinear set in Q6.

(i) $O(0,0)$O(0,0), $P(4,0)$P(4,0), $Q(4,4)$Q(4,4). $OP$OP is horizontal with length $4$4; $PQ$PQ is vertical with length $4$4 — equal legs meeting at a right angle at $P$P, so $OPQ$OPQ is a right-angled isosceles triangle (hypotenuse $OQ=4\sqrt2$OQ=4√2).

(ii) $O(0,0)$O(0,0), $A(-3,-3)$A(-3,-3) in Quadrant III, $B(3,-3)$B(3,-3) in Quadrant IV. $OA=\sqrt{9+9}=\sqrt{18}$OA=√(9+9)=√18 and $OB=\sqrt{9+9}=\sqrt{18}$OB=√(9+9)=√18 — equal, so $OAB$OAB is isosceles with one vertex ($A$A) in Quadrant III and one ($B$B) in Quadrant IV.

Compute the actual midpoint of $S$S and $T$T in each row and compare it with $M$M:

Row 1: midpoint of $ST=\left(\frac{-3+3}{2},\frac{0+0}{2}\right)=(0,0)=M$ST=((-3+3)/2,(0+0)/2)=(0,0)=M. Yes, M is the midpoint.

Row 2: midpoint of $ST=\left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4)=M$ST=((2+4)/2,(3+5)/2)=(3,4)=M. Yes, M is the midpoint.

Row 3: midpoint of $ST=\left(\frac{0+0}{2},\frac{0+(-10)}{2}\right)=(0,-5)$ST=((0+0)/2,(0+(-10))/2)=(0,-5), but $M=(0,5)\ne(0,-5)$M=(0,5)≠(0,-5). No, M is not the midpoint.

Row 4: midpoint of $ST=\left(\frac{-8+6}{2},\frac{7+(-3)}{2}\right)=(-1,2)$ST=((-8+6)/2,(7+(-3))/2)=(-1,2), but $M=(0,-2)\ne(-1,2)$M=(0,-2)≠(-1,2). No, M is not the midpoint.

Connection: whenever $M$M is the midpoint of $ST$ST, each coordinate of $M$M is the average of the corresponding coordinates of $S$S and $T$T: $M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$M=((x₁+x₂)/2,(y₁+y₂)/2) — the midpoint formula.

Using the midpoint formula with $A(3,-4)$A(3,-4), $M(-7,1)$M(-7,1): $-7=\dfrac{3+x}{2}\Rightarrow 3+x=-14\Rightarrow x=-17$-7=(3+x)/2⇒ 3+x=-14⇒ x=-17. $1=\dfrac{-4+y}{2}\Rightarrow -4+y=2\Rightarrow y=6$1=(-4+y)/2⇒ -4+y=2⇒ y=6. So $B=(-17,\,6)$B=(-17, 6).

Method: trisection points split $AB$AB into three equal parts, so $P$P is the point $\frac13$1/3 of the way from $A$A to $B$B, and $Q$Q is $\frac23$2/3 of the way — extending the averaging idea behind the midpoint formula (which finds the point "$\frac12$1/2 of the way"): $P=A+\frac13(B-A)=\dfrac{2A+B}{3}$P=A+1/3(B-A)=(2A+B)/3, and $Q=A+\frac23(B-A)=\dfrac{A+2B}{3}$Q=A+2/3(B-A)=(A+2B)/3.

For $A(4,7)$A(4,7), $B(16,-2)$B(16,-2): $B-A=(12,-9)$B-A=(12,-9).

$P=A+\frac13(12,-9)=(4+4,\,7-3)=(8,4)$P=A+1/3(12,-9)=(4+4, 7-3)=(8,4).

$Q=A+\frac23(12,-9)=(4+8,\,7-6)=(12,1)$Q=A+2/3(12,-9)=(4+8, 7-6)=(12,1).

So $P=(8,4)$P=(8,4) and $Q=(12,1)$Q=(12,1).

(i) Points lie on a common circle centred at $O$O exactly when they are all equidistant from $O$O.

$OA=\sqrt{1^2+(-8)^2}=\sqrt{1+64}=\sqrt{65}$OA=√(1²+(-8)²)=√(1+64)=√65

$OB=\sqrt{(-4)^2+7^2}=\sqrt{16+49}=\sqrt{65}$OB=√((-4)²+7²)=√(16+49)=√65

$OC=\sqrt{(-7)^2+(-4)^2}=\sqrt{49+16}=\sqrt{65}$OC=√((-7)²+(-4)²)=√(49+16)=√65

All three distances equal $\sqrt{65}$√65, so $A$A, $B$B, $C$C lie on a circle $K$K centred at $O$O with radius $\sqrt{65}$√65 units.

(ii) $OD=\sqrt{(-5)^2+6^2}=\sqrt{25+36}=\sqrt{61}\approx7.81$OD=√((-5)²+6²)=√(25+36)=√61≈7.81. Since $\sqrt{61}<\sqrt{65}$√61<√65, $D$D lies inside circle $K$K.

$OE=\sqrt{0^2+9^2}=\sqrt{81}=9$OE=√(0²+9²)=√81=9. Since $9>\sqrt{65}\approx8.06$9>√65≈8.06, $E$E lies outside circle $K$K.

Take $D$D as the midpoint of $BC$BC, $E$E as the midpoint of $CA$CA, $F$F as the midpoint of $AB$AB. Adding these: $D+E+F=\frac{(B+C)+(C+A)+(A+B)}{2}=A+B+C$D+E+F=(B+C)+(C+A)+(A+B)/2=A+B+C.

Also $E+F-D=\frac{(C+A)+(A+B)-(B+C)}{2}=\frac{2A}{2}=A$E+F-D=(C+A)+(A+B)-(B+C)/2=2A/2=A, and similarly $B=D+F-E$B=D+F-E, $C=D+E-F$C=D+E-F.

With $D(5,1)$D(5,1), $E(6,5)$E(6,5), $F(0,3)$F(0,3):

$A=E+F-D=(6+0-5,\;5+3-1)=(1,7)$A=E+F-D=(6+0-5, 5+3-1)=(1,7)

$B=D+F-E=(5+0-6,\;1+3-5)=(-1,-1)$B=D+F-E=(5+0-6, 1+3-5)=(-1,-1)

$C=D+E-F=(5+6-0,\;1+5-3)=(11,3)$C=D+E-F=(5+6-0, 1+5-3)=(11,3)

Check: midpoint of $B(-1,-1)$B(-1,-1) and $C(11,3)$C(11,3) is $(5,1)=D$(5,1)=D ✓; midpoint of $C(11,3)$C(11,3) and $A(1,7)$A(1,7) is $(6,5)=E$(6,5)=E ✓; midpoint of $A(1,7)$A(1,7) and $B(-1,-1)$B(-1,-1) is $(0,3)=F$(0,3)=F ✓.

So $A(1,7)$A(1,7), $B(-1,-1)$B(-1,-1), $C(11,3)$C(11,3).

(i) Draw a $10\times10$10×10 grid: 10 equally spaced vertical lines (the N–S streets) and 10 equally spaced horizontal lines (the E–W streets), each pair of adjacent parallel lines $1$1 cm apart on paper (representing the real $200$200 m spacing at the scale $1\text{ cm}=200\text{ m}$1 cm=200 m).

(ii) Each intersection label $(n,e)$(n,e) names a specific N–S street ($n$n) crossing a specific E–W street ($e$e) — since exactly one N–S street and exactly one E–W street carry any given index, each ordered pair names exactly one unique intersection.

(a) There is exactly 1 intersection labelled $(4,3)$(4,3) — where the 4th N–S street meets the 3rd E–W street.

(b) There is exactly 1 intersection labelled $(3,4)$(3,4) — where the 3rd N–S street meets the 4th E–W street. (This is a different crossing from $(4,3)$(4,3), since — just as with ordinary coordinates — the order of the two numbers matters.)

The screen occupies $0\le x\le800$0≤ x≤800, $0\le y\le600$0≤ y≤600.

Circle A (centre $(100,150)$(100,150), radius $80$80): extends from $x=100-80=20$x=100-80=20 to $x=100+80=180$x=100+80=180 (both within $[0,800]$[0,800]), and from $y=150-80=70$y=150-80=70 to $y=150+80=230$y=150+80=230 (both within $[0,600]$[0,600]). So circle A is entirely on-screen.

Circle B (centre $(250,230)$(250,230), radius $100$100): extends from $x=250-100=150$x=250-100=150 to $x=250+100=350$x=250+100=350, and $y=230-100=130$y=230-100=130 to $y=230+100=330$y=230+100=330 — all within the screen bounds. So circle B is also entirely on-screen.

(i) Neither circle lies partly outside the screen.

(ii) Distance between centres: $\sqrt{(250-100)^2+(230-150)^2}=\sqrt{150^2+80^2}=\sqrt{22500+6400}=\sqrt{28900}=170$√((250-100)²+(230-150)²)=√(150²+80²)=√(22500+6400)=√28900=170.

Sum of radii $=80+100=180$=80+100=180; difference of radii $=|100-80|=20$=|100-80|=20. Since $20<170<180$20<170<180 (the distance between centres is less than the sum of the radii but more than their difference), the two circles do intersect — they overlap, crossing at two points.

Sides:

$AB=\sqrt{(-1-2)^2+(2-1)^2}=\sqrt{9+1}=\sqrt{10}$AB=√((-1-2)²+(2-1)²)=√(9+1)=√10

$BC=\sqrt{(-2-(-1))^2+(-1-2)^2}=\sqrt{1+9}=\sqrt{10}$BC=√((-2-(-1))²+(-1-2)²)=√(1+9)=√10

$CD=\sqrt{(1-(-2))^2+(-2-(-1))^2}=\sqrt{9+1}=\sqrt{10}$CD=√((1-(-2))²+(-2-(-1))²)=√(9+1)=√10

$DA=\sqrt{(2-1)^2+(1-(-2))^2}=\sqrt{1+9}=\sqrt{10}$DA=√((2-1)²+(1-(-2))²)=√(1+9)=√10

All four sides are equal ($\sqrt{10}$√10 each), so $ABCD$ABCD is at least a rhombus. Now check the diagonals:

$AC=\sqrt{(-2-2)^2+(-1-1)^2}=\sqrt{16+4}=\sqrt{20}$AC=√((-2-2)²+(-1-1)²)=√(16+4)=√20

$BD=\sqrt{(1-(-1))^2+(-2-2)^2}=\sqrt{4+16}=\sqrt{20}$BD=√((1-(-1))²+(-2-2)²)=√(4+16)=√20

The diagonals are also equal ($\sqrt{20}$√20 each). A rhombus with equal diagonals must have right angles at its vertices, so $ABCD$ABCD is indeed a square.

Area $=\text{side}^2=(\sqrt{10})^2=10$=side²=(√10)²=10 square units. (Check via diagonal: area $=\dfrac{d^2}{2}=\dfrac{20}{2}=10$=d²/2=20/2=10 ✓.)